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Homework Help: Gauss's Law charge Problem

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data

    The box-like Gaussian surface of the attached figure encloses a net charge of [tex]+24.0 \epsilon_0 \mu C[/tex] and lies in an electric field given by

    [tex]\vec E = [(10.0 + 2.00x) \hat i - 2.00 \hat j + bz\hat k][/tex]N/C

    with x and z in meters and b a constant. The bottom face is in the xz plane; the top face is in the horizontal plane passing through y2 = 1.00 m. For x1 = 1.00 m, x2 = 4.00 m, z1 = 1.00 m, and z2 = 3.00 m, what is b?

    2. Relevant equations

    [tex]\Phi = \int{E dA}[/tex]

    [tex]\int{E dA} = \frac{Q}{\epsilon_0}[/tex]

    3. The attempt at a solution

    I get that some integration needs to take place over each surface and that the sum of each integration will equal 24 micro coulombs. I can then solve for b. But I'm not sure how the integration needs to happen. What variable do I integrate wrt both x and z? How does that work for the left or right side of the shape? Help very much appreciated!!
     

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  2. jcsd
  3. May 6, 2010 #2

    vela

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    You need six separate integrals, one for each face. The orientation of a face dictates which variables you integrate over. For example, for the righthand face, x is constant, so you'll integrate over y and z.
     
  4. May 6, 2010 #3
    Another hint (or caution) is to be careful how you write the integral equations for each surface. Strictly, the integral is a dot product of E and dA, so you will integrate the component of E which is perpendicular to the surface you are integrating over.
     
  5. May 6, 2010 #4
    So you're saying that I won't use i, j, k for all components? Only those applicable.

    For example, sticking with the right face, I would use all of them (as it's in i, j and and k) but for the bottom I'd only use i and k?

    Thanks.
     
  6. May 6, 2010 #5

    vela

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    Not quite. The vector dA points in the direction of the outward normal, so on the righthand face, for example, it would point in the +x direction. In other words, you'd have

    [tex]d\vec{A} = \hat{i}\, dA = \hat{i}\, dy\, dz[/tex]

    Now if you plug that into the integrand along with your expression for the electric field, what do you get for [tex]\vec{E}\cdot d\vec{A}[/tex]?
     
  7. May 6, 2010 #6
    No, that's not correct. We should back up one step before stating that we want to integrate the dot product of the field and the dA. We need to be clear about the direction of the area element. The area element dA can be thought of as having a direction outward and perpendicular to the surface. So the bottom surface has direction -j, for example. This means that the dot product with E only needs to consider the j-component of E on that surface.
     
  8. May 6, 2010 #7
    Would it be,

    [tex]\int^{z_1}_{z_2} \int^{y_2}_{0} -2.00 \hat j + bz \hat k dy dz[/tex]

    Sorry if I seem like a fish out of water.. but I actually haven't had multivariable calc... it isn't supposed to be required for the class ...
     
  9. May 6, 2010 #8

    vela

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    Since you're using vector notation, I assume you've learned some basic vector algebra. That's all you really need here. Recall that the dot product of two vectors is

    [tex]\vec{A}\cdot\vec{B} = A_x B_x + A_y B_y + A_z B_z[/tex]

    where [tex]\vec{A} = A_x\hat{i}+A_y\hat{j}+A_z\hat{k}[/tex] and [tex]\vec{B} = B_x\hat{i}+B_y\hat{j}+B_z\hat{k}[/tex]. Note that the dot product yields a number for an answer, not a vector.

    So try again.
     
  10. May 6, 2010 #9
    If I'm understanding correctly,

    [tex]\vec E \cdot d \vec A = 10.00 + 2.00x dy dz \hat i [/tex]

    Correct?
     
  11. May 6, 2010 #10

    vela

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    Not quite. (I'm not sure if you made a typo, but dA is a single quantity. It's not d times A.) You should have

    [tex]\vec{E}\cdot d\vec{A} = (10.00+2.00x)\,dy\,dz[/tex]

    The answer is not a vector, so the i-hat you had shouldn't be there.
     
  12. May 6, 2010 #11
    So... I need to determine [tex]d \vec A[/tex] and dot it with [tex] \vec E[/tex] for each face of the shape. Integrate to remove the differentials, then sum the result of each integration to solve for b?

    Am I finally on the right track?

    Thank you so much for your help so far.
     
  13. May 6, 2010 #12

    vela

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    Yeah, that's exactly what you want to do.
     
  14. May 6, 2010 #13
    Something isn't working out. Each side cancels the other side... including the sides which involve b.

    Looking at the left face,

    [tex]d \vec A = -i dy dz[/tex]

    and

    [tex] \vec E \cdot d \vec A = -(10.00 + 2.00x) dy dz[/tex]

    Integrating the left and right face each over [tex] z_1 [/tex] to [tex] z_2 [/tex] and 0 to [tex] y_2[/tex] gives 48 and -48 respectively.

    What am I doing wrong?
     
  15. May 6, 2010 #14

    vela

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    You're not integrating correctly. Remember x is a constant on those two faces, so you can just plug in what it equals for each face. You should get 36 and -24 for the two integrals (if I didn't make a mistake doing it in my head).
     
  16. May 6, 2010 #15
  17. May 6, 2010 #16

    vela

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    Where did the 3 come from in those integrals? Are you sure that's right for both of them?
     
  18. May 6, 2010 #17
    [tex]z_2 = 3.00 meters[/tex]
     
  19. May 6, 2010 #18

    vela

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    You indeed have z=z2 for the front face, but what about the back face of the block?
     
  20. May 7, 2010 #19
    That was the problem.. I was using the wrong values for the constants.... I'll work it out and let you know if I have any more problems. Thanks for the help!
     
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